Thread: Derivatives using ln

Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)
= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.

Originally Posted by Jools

Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)


From where did you bring that ? It can only confuse you.

= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)


Here's the mistake: you actually have , which is positive indeed.

Tonio


In fact now I realized that that -2 is a power (!!), but still: you wrote , which is lacking a minus sign there!

Tonio


= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.

.

Originally Posted by Jools

Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)
= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.

does not equal

Originally Posted by Jools

Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'

500(-1)(1+e^-t)^-2 (-(e^-t)) = 500 (1+e^-t)^-2 (e^-t)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6
Since the derivative is 500(1 + e^-3)^-2(e^-3), your value should not come up negative now
I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.

The derivative of is

Would I also be correct in writing it like this:

-500(1 + e^-t)^-2 (-t e^-t-1)

Then solving for P '(3)?

It doesn't give the same answer when I solve for P '(3) as I get with your method, so I am assuming no...